Question

Adsorption of a gas follows Freundlich adsorption isotherm
$\frac{x}{m}=k{P}^{\frac{1}{n}}$

$x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent

The plot of $\log \left(\frac{x}{m}\right)$ vs $\log P$ is shown,
The value of n is:

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{1}{12}$
• D. $12$

Answer 1

The correct option is A $\frac{4}{3}$

Freundlich adsorption isotherm:
$\frac{x}{m}=k{P}^{\frac{1}{n}}..\left(i\right)$
Taking log both sides
$\log \left(\frac{x}{m}\right)=\log \left(k{P}^{\frac{1}{n}}\right)$
$\log \left(\frac{x}{m}\right)=\log \left(k\right)+\frac{1}{n}\log P$
The slope is : $\frac{1}{n}$
From graph:
$\text{Slope}=\frac{3}{4}=\frac{1}{n}$
$n=\frac{4}{3}$